Figure this out as though no digits are present. You can choose 8 of the 26 letters and that's it.
26^8 [you can allow repeating letters] = 2.088 * 10^11
The total number of ways that can be chosen with letters and digits intermixed = 36^8
The probability of getting no numbers is 26^8 / 36^8 or (26/36)^8 = 0.07402
So the probability that there were be at least one digit is
1 - 0.07402 = 0.92598
Answer:
(- 2, 4 )
Step-by-step explanation:
Given endpoints (x₁, y₁ ) and (x₂, y₂ ) , then the midpoint is
(
,
)
Here (x₁, y₁ ) = A (4, 6 ) and (x₂, y₂ ) = B (- 8, 2 )
midpoint = (
,
) = (
,
) = (- 2, 4 )
Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.
Step-by-step explanation:
Total number of students = 8
Number of student who has passed Exam P/1 = 1
Number of student who has passed Exam FM/2 = 1
No student has passed more than one exam.
According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.
So, Probability will be

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.
What is this question asking?
maybe try 3.87